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IV I.Machines

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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
IV. Three-Phase Induction Machines
Induction Machines
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
Example 1:
A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor has the following parameters:
R 1 =0.461 Ω, R 2 =0.258 Ω, X 1 =0.507 Ω, X 2 =0.309 Ω, X m =30.74 Ω
Rotational losses are 2450W. The motor drives a mechanical load at a speed of 1170 rpm. Calculate
the following information:
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
Synchronous speed in rpm
slip
Line Current
Input Power
Airgap Power
Torque Developed
Output Power in Hp
Efficiency
This machine has no iron loss resistance, so the equivalent circuit is as follows:
i.
Synchronous speed is given by:
ii. Slip is given by
Using the rpm equation,
s = (1200-1170)/1200 = 0.025
iii. Now, phase current is given by
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
where phase impedance is given by
Using the above equation, Z in = 9.57 + j3.84 Ω
And noting that the machine is delta connected, V 1 = V LL = 480V
I 1 = 43.1 - j17.4 A. |I 1 | =46.6 A, θ = -21.9°
Therefore I L = √3 × 46.6 = 80.6 A
iv. Input power is given by:
P in = 62.2 kW
v.
Airgap power is the input power minus stator losses. In this case the core losses are grouped
with rotational loss. Therefore
P gap = 62.2 kW - 3× 46.62 × 0.461
P gap = 59.2 kW
vi.
Torque developed can be found from
where synchronous speed in radians per second is given by
giving
τ = 471 Nm
Vii.Output power in horsepower is the output power in Watts divided by 746. (there are 746 W in one Hp).
and
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
Therefore output power in Watts is: P out = 55.3kW
Viii. Efficiency is given by
η = 55.3/62.2 = 88.9%
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
Using the induction machine power and torque equations it is possible to produce the
torque speed curve shown below.
Operating Regions
The torque-speed curve brakes down into three operating regions:
1. Braking, n m < 0, s > 1
Torque is positive whilst speed is negative. Considering the power conversion
equation
it can be seen that if the power converted is negative (from P = τ ω) then the
airgap power is positive. i.e. the power is flowing from the stator to the rotor and
also into the rotor from the mechanical system. This operation is also called
plugging.
This mode of operation can be used to quickly stop a machine. If a motor is
travelling forwards it can be stopped by interchanging the connections to two of
the three phases. Switching two phases has the result of changing the direction of
motion of the stator magnetic field, effectively putting the machine into braking
mode in the opposite direction.
2. Motoring, 0 < n m < n s , 1 > s > 0
Torque and motion are in the same direction. This is the most common mode of
operation.
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
3. Generating, n m > n s , s < 0
In this mode, again torque is positive whilst speed is negative. However, unlike
plugging,
indicates that if the power converted is negative, so is the air gap power. In this
case, power flows from the mechanical system, to the rotor circuit, then across the
air gap to the stator circuit and external electrical system.
Motoring Torque Characteristic
The motoring region of the induction machine torque-speed curve is the region of greatest interest.
• Starting torque
– Torque at zero speed
– Typically 1.5 times the full-load torque
• Pull-up torque
– The minimum torque developed by the motor while accelerating from zero speed
– Greater than full-load torque; less than starting torque
• Breakdown torque
– The maximum torque that the motor can develop
– Typically 2.5 times the full-load torque
• Normal operation
– At full-load, the motor runs at n rpm
– Rotor speed decreases slightly from synchronous speed with increasing load torque
– Motor will stall when the load torque exceeds the breakdown torque
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
The torque equation
Using the equation
Multiple solutions of the above equation for torque at different slips can be made simpler by
simplifying the equivalent circuit model. Consider the diagram below:
The stator part of the equivalent circuit (together with the magnetising branch) can be replaced by a
Thevenin equivalent circuit. In the Thevenin circuit, the stator phase voltage has been replaced by its
Thevenin equivalent,
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
and the impedances have been replaced by Thevenin equivalent impedances.
Incorporating the Thevenin model into the circuit model results in the Thevenin equivalent circuit
model of an induction machine.
In the above circuit, the calculation of rotor current is greatly simplified
The above expression for rotor current can be squared and substituted into the torque equation
Using the above equation, the variation of torque with slip can be plotted directly .
Note that if power or efficiency calculations are needed, the full equivalent circuit model should
be used (not the Thevenin version).
since synchonous speed is constant, maximum torque occurs at the same slip as maximum airgap power.
Considering the Thevenin circuit, and applying maximum power transfer theory, maximum airgap power and
maximum torque will occur when
𝑅𝑅2
2
= �𝑅𝑅𝑇𝑇𝑇𝑇
+ (𝑋𝑋𝑇𝑇𝑇𝑇 + 𝑋𝑋2 )2
𝑠𝑠
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
Re-arranging it is possible to obtain the slip for maxiumum torque, or pullout torque.
π‘†π‘†π‘šπ‘šπ‘šπ‘šπ‘šπ‘š =
𝑅𝑅2
2
�𝑅𝑅𝑇𝑇𝑇𝑇
+ (𝑋𝑋𝑇𝑇𝑇𝑇 + 𝑋𝑋2 )2
Substituting the pullout slip into the Thevenin torque equation:
π‘‡π‘‡π‘šπ‘šπ‘šπ‘šπ‘šπ‘š =
2
3𝑉𝑉𝑇𝑇𝑇𝑇
2πœ”πœ”π‘ π‘ π‘ π‘ π‘ π‘ π‘ π‘  �𝑅𝑅𝑇𝑇𝑇𝑇 + �𝑅𝑅2𝑇𝑇𝑇𝑇 + (𝑋𝑋𝑇𝑇𝑇𝑇 + 𝑋𝑋2 )2 οΏ½
The Starting torque
•
•
•
•
2
The starting torque is proportional to 𝑉𝑉𝑇𝑇𝑇𝑇
𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = πœ”πœ”
2
3𝑉𝑉𝑇𝑇𝑇𝑇
𝑅𝑅2
2 +(𝑋𝑋
2
[(𝑅𝑅
)
+𝑅𝑅
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑇𝑇𝑇𝑇
2
𝑇𝑇𝑇𝑇 +𝑋𝑋2 ) ]
The starting torque will be reduced if the stator and rotor leakage inductances are increased
The starting torque will be reduced if the stator frequency is increased
When the rotor resistance is increased, the starting torque will first increase and then decrease.
𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 π‘šπ‘šπ‘šπ‘šπ‘šπ‘š = π‘‡π‘‡π‘šπ‘šπ‘šπ‘šπ‘šπ‘š when 𝑅𝑅2= 𝑋𝑋1 + 𝑋𝑋2
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
Equivalent Circuit Model Analysis Example
A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor has the following parameters:
R 1 =0.461 Ω, R 2 =0.258 Ω, X 1 =0.507 Ω, X 2 =0.309 Ω, X m =30.74 Ω
Rotational losses are 2450W. The motor drives a mechanical load at a speed of 1170 rpm. Find:
i.
ii.
iii.
iv.
Thevenin circuit parameters and Thevenin voltage
Pullout slip
Pullout Torque
Start Torque
Solution:
Using Matlab or Excel (or another computer program) plot the torque speed curve for slip in the range
0 to 1
i.
Thevenin circuit parameters and Thevenin voltage:
The Thevenin voltage is the voltage applied to the rotor assuming that the rotor current is
zero. Thevenin impedance is the impedance of the stator part of the circuit, seen from the rotor,
assuming that the stator supply is short circuited.
i.
Substituting the equivalent circuit parameters in to the above equations gives:
V TH = 475.2 V, R TH = 0.452Ω, X TH = 0.313Ω
i.
Pullout slip
The slip at which maximum torque occurs can be found from maximum power transfer theory.
Maximum torque and maximum airgap power occur at the same slip, therefore maximum
torque occurs when
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IV. Three-phase Induction Machines
ii.
Dr. Suad Ibrahim Shahl
Pullout Torque
Pullout torque can be found by substituting the above pullout slip into the Thevenin torque
equation
or from the maximum torque equation directly
Substituting into the above equation:
i.
Start Torque
Start torque can be found by setting s=1 in the above equation for torque.
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
Effect of Rotor Resistance
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
Investigating the torque-speed curve it is apparent that the rotor circuit resistance has significant
impact on speed at which maximum torque occurs. The plots below illustrate two cases, with low
rotor resistance on the left and high rotor resistance on the right.
Example 1: A 460-V, 25-hp, 60-Hz, four-pole, Y-connected wound-rotor induction motor has the following
impedances in ohms per phase referred to the stator circuit:
π‘Ήπ‘ΉπŸπŸ = 𝟎𝟎. πŸ”πŸ”πŸ”πŸ”πŸ”πŸ”π›€π›€
π‘Ώπ‘ΏπŸπŸ = 𝟏𝟏. πŸπŸπŸπŸπŸπŸπ›€π›€
π‘Ήπ‘ΉπŸπŸ = 𝟎𝟎. πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘π›€π›€
π‘Ώπ‘ΏπŸπŸ = 𝟎𝟎. πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’π›€π›€
𝑿𝑿𝑴𝑴 = 𝟐𝟐𝟐𝟐. πŸ‘πŸ‘π›€π›€
a) What is the maximum torque of this motor? At what speed and slip does it occur?
b) What is the starting torque of this motor?
c) When the rotor resistance is doubled, what is the speed at which the maximum torque now occurs?
What is the new starting torque of the motor?
Solution: The Thevenin voltage of this motor is
The Thevenin resistance is
The Thevenin reactance is
𝑉𝑉𝑇𝑇𝑇𝑇 = π‘‰π‘‰πœ™πœ™
=
2
𝑋𝑋𝑀𝑀
οΏ½
1 +𝑋𝑋 𝑀𝑀
𝑅𝑅𝑇𝑇𝑇𝑇 ≈ 𝑅𝑅1 οΏ½
𝑋𝑋
π‘‹π‘‹π‘‡π‘‡β„Ž ≈ 𝑋𝑋1 ≈ 1.106𝛀𝛀
�𝑅𝑅12 +(𝑋𝑋1 +𝑋𝑋𝑀𝑀 )2
(266 𝑉𝑉)(26.3 𝛀𝛀 )
οΏ½(𝟎𝟎.πŸ”πŸ”πŸ”πŸ”πŸ”πŸ”π›€π›€ )2 +(𝟏𝟏.πŸπŸπŸπŸπŸπŸπ›€π›€ +𝟐𝟐𝟐𝟐.πŸ‘πŸ‘π›€π›€ )2
2
𝟐𝟐𝟐𝟐.πŸ‘πŸ‘π›€π›€
οΏ½
𝟏𝟏.πŸπŸπŸπŸπŸπŸπ›€π›€+𝟐𝟐𝟐𝟐.πŸ‘πŸ‘π›€π›€
≈ (𝟎𝟎. πŸ”πŸ”πŸ”πŸ”πŸ”πŸ”π›€π›€) οΏ½
𝑋𝑋𝑀𝑀
= 255.2𝑉𝑉
= 0.590𝛀𝛀
(a) The slip at which maximum torque occurs is given by Equation (2):
π‘†π‘†π‘šπ‘šπ‘šπ‘šπ‘šπ‘š =
𝑅𝑅2
2 +(𝑋𝑋
2
�𝑅𝑅𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇 +𝑋𝑋2 )
=
𝟎𝟎.πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘π›€π›€
οΏ½(0.590𝛀𝛀 )2 +(1.106𝛀𝛀+𝟎𝟎.πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’π›€π›€ )2
= 0.198
This corresponds to a mechanical speed of
π‘›π‘›π‘šπ‘š = (1 − 𝑠𝑠)𝑛𝑛𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 β„Ž = (1 − 0.198)(1800 π‘Ÿπ‘Ÿ/π‘šπ‘šπ‘šπ‘šπ‘šπ‘š) = 1444 π‘Ÿπ‘Ÿ⁄π‘šπ‘šπ‘šπ‘šπ‘šπ‘š
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
The torque at this speed is
π‘‡π‘‡π‘šπ‘šπ‘šπ‘šπ‘šπ‘š =
2
3𝑉𝑉𝑇𝑇𝑇𝑇
2 +(𝑋𝑋
2
2πœ”πœ” 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �𝑅𝑅𝑇𝑇𝑇𝑇 +�𝑅𝑅𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇 +𝑋𝑋2 ) οΏ½
3(255.2 𝑉𝑉)2
= 2(188.5 π‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿ /𝑠𝑠)[(0.590𝛀𝛀)2 +(1.106𝛀𝛀+𝟎𝟎.πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’)2 ] = 229 𝑁𝑁. π‘šπ‘š
(b) The starting torque of this motor is found by setting s=1 in Equation (1) :
𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
=
2
3𝑉𝑉𝑇𝑇𝑇𝑇
𝑅𝑅2
2
πœ”πœ”π‘ π‘ π‘ π‘ π‘ π‘ π‘ π‘  [(𝑅𝑅𝑇𝑇𝐻𝐻 + 𝑅𝑅2 ) + (𝑋𝑋𝑇𝑇𝑇𝑇 + 𝑋𝑋2 )2 ]
3(255.2 𝑉𝑉)2 (𝟎𝟎. πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘π›€π›€)
= 104 𝑁𝑁. π‘šπ‘š
(188.5 π‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿ⁄𝑠𝑠)[(0.590𝛀𝛀 + 𝟎𝟎. πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘π›€π›€)2 + (1.106𝛀𝛀 + 𝟎𝟎. πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’)2 ]
(c ) If the rotor resistance is doubled, then the slip at maximum torque doubles, too. Therefore,
π‘ π‘ π‘šπ‘šπ‘šπ‘šπ‘šπ‘š = 0.396
and the speed at maximum torque is
π‘›π‘›π‘šπ‘š = (1 − 𝑠𝑠)𝑛𝑛𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = (1 − 0.396)(1800 π‘Ÿπ‘Ÿ⁄π‘šπ‘šπ‘šπ‘šπ‘šπ‘š) = 1087 π‘Ÿπ‘Ÿ⁄π‘šπ‘šπ‘šπ‘šπ‘šπ‘š
The maximum torque is still
π‘‡π‘‡π‘šπ‘šπ‘šπ‘šπ‘šπ‘š = 229 𝑁𝑁. π‘šπ‘š
The starting torque is now
𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
3(255.2𝑉𝑉)2 (0.664𝛀𝛀)
= 170𝑁𝑁. π‘šπ‘š
(188.5 π‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘‘π‘‘⁄𝑠𝑠)[(0.590𝛀𝛀 + 𝟎𝟎. πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’)2 + (1.106𝛀𝛀 + 𝟎𝟎. πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’)2 ]
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
When the switch S is in the START position, the stator windings are connected in STAR
When the motor picks up speed, say 80% of its rated value, the switch S is thrown quickly to the RUN
position which connects the stator windings in DELTA.
In induction motors that are designed to operate with delta stator connection it is possible, during starting, to
reduce the phase voltage by switching to Y- connection
During Y- connection, the phase voltage V s becomes
so the phase current, for same slip, IsY , is reduced 3 times
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
Now the line current in Δ connection I lΔ is
So the line current is three times smaller for Y- connection. The torque is proportional to phase voltage squared
Therefore, the Y-delta starting is equivalent to
in torque.
a reduction of phase voltage and a 3 to 1 reduction
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
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Dr. Suad Ibrahim Shahl
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Fig.1 Slip ring IM
Fig.2 Torque-slip curves
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IV. Three-phase Induction Machines
Dr. Suad Ibrahim Shahl
Fig.3 Torque-speed curves
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